題目
題目描述
大家應該都會玩“錘子剪刀布”的游戲：
現給出兩人的交鋒記錄，請統計雙方的勝、平、負次數，并且給出雙方分別出什么手勢的勝算最大。

輸入描述:
輸入第1行給出正整數N（<=105），即雙方交鋒的次數。隨后N行，每行給出一次交鋒的信息，即甲、乙雙方同時給出的的手勢。C代表“錘子”、J代表“剪刀”、B代
表“布”，第1個字母代表甲方，第2個代表乙方，中間有1個空格。

輸出描述:
輸出第1、2行分別給出甲、乙的勝、平、負次數，數字間以1個空格分隔。第3行給出兩個字母，分別代表甲、乙獲勝次數最多的手勢，中間有1個空格。如果解不唯
一，則輸出按字母序最小的解。

輸入例子:
10
C J
J B
C B
B B
B C
C C
C B
J B
B C
J J

輸出例子:
5 3 2
2 3 5
B B

代碼
第一次用C++做題，感受到了面向對象思想的妙處

#define _CRT_SECURE_NO_WARNINGS #include<iostream> #include<stdlib.h> int i; class People { public:int cur = 0;int win = 0;int loose = 0;int equal = 0;int winbecauseof_J = 0;int winbecauseof_B = 0;int winbecauseof_C = 0;char a[105] = { 0 };//最多容納字母個數int play(int total); }; class Checker { public:int cur_ckr = 0;//正在檢查的局數int check(People *jia, People *yi, int total);//參數：對象甲，對象乙，需要檢查的局數 }; int People::play(int cur) {// std::cout << "請輸入第" << cur+1 << "個字母"<<std::endl;std::cin >> this->a[cur];cur++;return cur; } //判斷勝負 int Checker::check(People *jia, People *yi, int total) {for (cur_ckr = 0; cur_ckr < total; cur_ckr++){if (jia->a[cur_ckr] == yi->a[cur_ckr])//平局{jia->equal++;yi->equal++;}else if ((jia->a[cur_ckr] == 'J'&& yi->a[cur_ckr] == 'B') || (jia->a[cur_ckr] == 'B'&& yi->a[cur_ckr] == 'C') || (jia->a[cur_ckr] == 'C'&& yi->a[cur_ckr] == 'J'))//甲勝利{jia->win++;yi->loose++;switch (jia->a[cur_ckr])//因為哪個字母而贏{case'J':jia->winbecauseof_J++; break;case'B':jia->winbecauseof_B++; break;case'C':jia->winbecauseof_C++; break;}}else if ((jia->a[cur_ckr] == 'B'&& yi->a[cur_ckr] == 'J') || (jia->a[cur_ckr] == 'C'&& yi->a[cur_ckr] == 'B') || (jia->a[cur_ckr] == 'J'&& yi->a[cur_ckr] == 'C'))//乙勝利{jia->loose++;yi->win++;switch (yi->a[cur_ckr])//因為哪個字母而贏{case'J':yi->winbecauseof_J++; break;case'B':yi->winbecauseof_B++; break;case'C':yi->winbecauseof_C++; break;}}else return -1;//返回錯誤代碼}return 0; } //打印 class Printer { public:void print_num(People *jia, People *yi){std::cout << jia->win << " " << jia->equal << " " << jia->loose << std::endl;std::cout << yi->win << " " << yi->equal << " " << yi->loose << std::endl;}void print_char(People *jia, People *yi){//甲if (jia->winbecauseof_J > jia->winbecauseof_C&& jia->winbecauseof_J > jia->winbecauseof_B){std::cout << "J ";}else if (jia->winbecauseof_C > jia->winbecauseof_J&& jia->winbecauseof_C > jia->winbecauseof_B){std::cout << "C ";}else if (jia->winbecauseof_B > jia->winbecauseof_C&& jia->winbecauseof_B > jia->winbecauseof_J){std::cout << "B ";}else if (jia->winbecauseof_J == jia->winbecauseof_B == jia->winbecauseof_C){std::cout << "B ";}else if (jia->winbecauseof_J == jia->winbecauseof_B || jia->winbecauseof_B == jia->winbecauseof_C){std::cout << "B ";}else if (jia->winbecauseof_J == jia->winbecauseof_C){std::cout << "C ";}//乙if (yi->winbecauseof_J > yi->winbecauseof_C&& yi->winbecauseof_J > yi->winbecauseof_B){std::cout << "J";}else if (yi->winbecauseof_C > yi->winbecauseof_J&& yi->winbecauseof_C > yi->winbecauseof_B){std::cout << "C";}else if (yi->winbecauseof_B > yi->winbecauseof_C&& yi->winbecauseof_B > yi->winbecauseof_J){std::cout << "B";}else if (yi->winbecauseof_J == yi->winbecauseof_B == yi->winbecauseof_C){std::cout << "B";}else if (yi->winbecauseof_J == yi->winbecauseof_B || yi->winbecauseof_B == yi->winbecauseof_C){std::cout << "B";}else if (yi->winbecauseof_J == yi->winbecauseof_C){std::cout << "C";}std::cout << std::endl;} }; int main() {int total;//總局數std::cin >> total;int checkok;People jia;People yi;Checker checker;//輸入每一局的情況(偶數輸入甲，奇數輸入乙)for (i = 0; i < 2 * total; i++){if (i % 2 == 0){jia.cur = jia.play(jia.cur);}else yi.cur = yi.play(yi.cur);}//逐個字母比較checkok = checker.check(&jia, &yi, total);//優化：定義為虛函數if (checkok == -1)//程序的健壯性{std::cout << "Oh NO! checker出現一點小問題,原因可能是你輸入的字母不對" << std::endl;}//打印Printer printer;printer.print_num(&jia, &yi);printer.print_char(&jia, &yi);system("pause"); } 超強干貨來襲 云風專訪：近40年碼齡，通宵達旦的技術人生

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